Question

Which aqueous solution has minimum freezing point?

• A. $0.01\mathrm{M}\mathrm{NaCl}$
• B. $0.005\mathrm{M}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$
• C. $0.005\mathrm{M}{\mathrm{MgI}}_2$
• D. $0.005\mathrm{M}{\mathrm{MgSO}}_4$

Answer 1

The correct option is A

$0.01\mathrm{M}\mathrm{NaCl}$

Explanation for correct answer:-

Option (A) $0.01\mathrm{M}\mathrm{NaCl}$

Step 1: Identifying the formula

We know that, $∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\times \mathrm{i}$

Where $∆{\mathrm{T}}_{\mathrm{f}}=$depression in freezing point, ${\mathrm{K}}_{\mathrm{f}}=$cryoscopic constant, $\mathrm{m}=$molality, $\mathrm{i}=$Van't Hoff factor

And we also that more the value of $∆{\mathrm{T}}_{\mathrm{f}}$ less will be the freezing point.

Step 2: Calculating the value of $∆{\mathrm{T}}_{\mathrm{f}}$

It is given that $\mathrm{m}=0.01$, and here $\mathrm{i}=2$

$$\begin{gathered} \therefore ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\times \mathrm{i} \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=1.86\times 0.01\times 2 \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=0.0372 \end{gathered}$$

Thus it's the maximum among the others.

Therefore, it has a minimum freezing point.

Hence it is the correct option.

The explanation for correct answer:-

Option (B) $0.005\mathrm{M}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

Step 1: Calculating the value of $∆{\mathrm{T}}_{\mathrm{f}}$

It is given that $\mathrm{m}=0.005$, and here $\mathrm{i}=1$

$$\begin{gathered} \therefore ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\times \mathrm{i} \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=1.86\times 0.005\times 1 \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=0.0093 \end{gathered}$$

This is not the maximum, thus the freezing point is not the minimum.

Hence it is the incorrect option.

Option (C) $0.005\mathrm{M}{\mathrm{MgI}}_2$

Step 1: Calculating the value of $∆{\mathrm{T}}_{\mathrm{f}}$

It is given that $\mathrm{m}=0.005$, and here $\mathrm{i}=3$

$$\begin{gathered} \therefore ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\times \mathrm{i} \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=1.86\times 0.005\times 3 \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=0.0279 \end{gathered}$$

This is not the maximum, thus the freezing point is not the minimum.

Hence it is the incorrect option.

Option (D) $0.005\mathrm{M}{\mathrm{MgSO}}_4$

Step 1: Calculating the value of $∆{\mathrm{T}}_{\mathrm{f}}$

It is given that $\mathrm{m}=0.005$, and here $\mathrm{i}=2$

$$\begin{gathered} \therefore ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\times \mathrm{i} \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=1.86\times 0.005\times 2 \\ \Rightarrow ∆{\mathrm{T}}_{\mathrm{f}}=0.0186 \end{gathered}$$

This is not the maximum, thus the freezing point is not the minimum.

Hence it is the incorrect option

Therefore, Option (A) $0.01\mathrm{M}\mathrm{NaCl}$ is the correct answer.